You start with 20 g of ice (H2O) at 260 K in a thermally isolated container. If 15500 J is added by heat, what will be the final temperature and state of the H2O? Show your work.
(Cice = 2.1 J/gK, Cwater = 4.18 J/gK, Cwatervapor = 1.7 J/gK, ΔHfus(H2O) = 334 J/g,
ΔHvap (H2O) = 2261 J/g, Melting Point = 273 K, Boiling Point = 373 K)
I'll get you started. K = 260 so that is -13 C.
How much heat is required to move the ice from -13 to zero? That's
q1 = mass ice x specific heat ice x (Tfinal - Tinitial) = 20 x 2.1 x [0 - (-13)] = ? J.
q2 to melt ice is mass ice x heat fusion = 20 x 2.1 = ? J.
q3 to move liquid water from zero C to 100 C = formula for q1
q4 to boil water = formula for q2 but heat vap
Starting with 15,500 J I would subtract q1, then q2, then q3 and continue until you have used almost all of the 15,500 and the next subtraction leaves you with a negative number which, of course, can't be. Then use the next formula in the series above to determine final T. Post your work if you get stuck.
To solve this problem, we need to consider the heat transfer that occurs during the different phases of water and the temperature changes associated with them.
1. Heat transfer during the phase change from ice to liquid water (melting):
The heat transfer during melting can be calculated using the formula:
Q = m * ΔHfus
where Q is the heat transferred, m is the mass of the substance, and ΔHfus is the heat of fusion.
In this case, the mass of ice is given as 20 g. Therefore, the heat transfer during melting is:
Q_melting = 20 g * 334 J/g = 6680 J
2. Heat transfer during the temperature change from 0°C (273 K) to the boiling point (373 K) of water:
The heat transfer during a temperature change can be calculated using the formula:
Q = m * C * ΔT
where Q is the heat transferred, m is the mass of the substance, C is the specific heat capacity, and ΔT is the change in temperature.
We need to consider the heat transfer during temperature change in two steps: from ice to 0°C and from 0°C to the boiling point.
a. Heat transfer from -13°C (260 K) to 0°C (273 K):
Q_ice_to_0 = 20 g * 2.1 J/gK * (0°C - (-13°C)) = 20 g * 2.1 J/gK * 13 K = 546 J
b. Heat transfer from 0°C (273 K) to the boiling point (373 K):
Q_0_to_boiling = 20 g * 4.18 J/gK * (373 K - 273 K) = 20 g * 4.18 J/gK * 100 K = 8360 J
3. Heat remaining for phase change from liquid water to water vapor (boiling):
The remaining heat available for the water to change from liquid to vapor is the difference between the total heat added and the heat transferred during temperature change and melting:
Q_remaining = 15500 J - Q_melting - Q_ice_to_0 - Q_0_to_boiling
Now, let's calculate the remaining heat:
Q_remaining = 15500 J - 6680 J - 546 J - 8360 J = 0 J
Since the remaining heat is zero, this means that there is no heat left for the water to undergo vaporization. Therefore, the final state of the H2O will be liquid water.
To find the final temperature, we need to calculate the temperature change from the boiling point to the final equilibrium temperature:
Q_final_temperature = m * C_water * (T_final - T_boiling)
0 J = 20 g * 4.18 J/gK * (T_final - 373 K)
T_final - 373 K = 0
T_final = 373 K
Hence, the final temperature of the water will be 373 K, which is the boiling point. The final state of the H2O will be liquid water.